Problem: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 6}{x - 8} = \dfrac{9x - 2}{x - 8}$
Solution: Multiply both sides by $x - 8$ $ \dfrac{x^2 + 6}{x - 8} (x - 8) = \dfrac{9x - 2}{x - 8} (x - 8)$ $ x^2 + 6 = 9x - 2$ Subtract $9x - 2$ from both sides: $ x^2 + 6 - (9x - 2) = 9x - 2 - (9x - 2)$ $ x^2 + 6 - 9x + 2 = 0$ $ x^2 + 8 - 9x = 0$ Factor the expression: $ (x - 8)(x - 1) = 0$ Therefore $x = 8$ or $x = 1$ However, the original expression is undefined when $x = 8$. Therefore, the only solution is $x = 1$.